If 321\(_4\) is divided by 23\(_4\) and leaves a remainder r, what is the value of r?
\(\frac{321_4}{23_4}\) = \(\frac{(3\times4^{2})+(2\times4^{1})+(1\times4^{0})}{(2\times4^{0})+(3\times4^{0})}\)
= \(\frac{3\times16+2\times4+1\times1}{2\times4+3\times1}\) = \(\frac{48+8+1}{8+3}\) =\(\frac{57}{11}\) = 5\(\hspace{1mm}remainder\hspace{1mm}2\)
∴ r = 2\(_{10}\) Now\(\hspace{1mm}convert\hspace{1mm}2_{10} \hspace{1mm}to\hspace{1mm}base\hspace{1mm}4\frac{4}{2}\) = 2\(\frac{4}{0}\) = 0\(\hspace{1mm}or\hspace{1mm}2\) ∴ r = 2
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